Differentials

So, let's say you have this tangent line: y - f(c) = f'(c)(x-c). In that equation, you want to find out (x, f(x)) and you're given (c, f(c)). So, x-c is the change in x, or Δx. The change in y, or Δy is y - f(c). So, you can rewrite the equation as Δy = f'(c)Δx. Usually, this isn't written with Δs. Instead, it's usually written as dy = f'(x)dx, and this is because dx is explicitly nonzero. Also, dy is an approximation, and Δy is the actual value. Note that this formula assumes that a function is locally linear, meaning that as two points get closer and closer together, the function becomes very much like a line.

It should also be noted that dy/dx = f'(x), which are equivalent notations of taking a derivative. This works because the derivative is just the instantaneous slope, which is Δy/Δx. If you know for sure that Δx != 0, then you can multiply by dx.

dy = f'(x)dx = (2x)(0.01) = 0.02x = 0.02(1) = 0.02

Δy = f(1.01) - f(1) = 1.0201 - 1 = 0.0201.

So, they ended up being very close, and when you're dealing with a very small values of dx, differentials become more and more accurate.

Estimating Errors

If you're measuring a square in cm, and you know the measurement is accurate to within 5 mm, or .5 cm. So, the measurement is x cm +- .5 cm. The error is affected differently though. The propagated error in that is (x +/- .5)^2 - x^2 = Δy = x +/- 0.25.

Using Differentials

Let's say the radius of a ball is supposed to be 1 inch, and your measurement is correct to within 0.01 inch, estimate the propagated error in the volume of the ball bearing.

Firstly, we let x = 1, and Δx = +/-0.01.

So, ΔV ≈ dV, dV = 4πx2(dx)

(4πx2 is the derivative of the volume of a sphere)

dx is +/- 0.01, and x = 1, so dV = 4π(1)^2 * (+/- 0.01)

dV = 0.1256 in3

That's the propagated error.

In order to find out whether the propagated error is big or small, you divide by the area.
In this case, you determine that by divided the propagated error by the actual value.